second total derivative of 3 variables

Question

I have a function: $$r=\sqrt{x^2+y^2+z^2}$$ and wish to calculate: $$\frac{d^2r}{dt^2}$$

---

so far I have said: $$\frac{d^2r}{dt^2}=\frac{\partial^2r}{\partial x^2}\left(\frac{dx}{dt}\right)^2+\frac{\partial^2r}{\partial y^2}\left(\frac{dy}{dt}\right)^2+\frac{\partial^2r}{\partial z^2}\left(\frac{dz}{dt}\right)^2+2\left[\frac{\partial^2r}{\partial x\partial y}\frac{dx}{dt}\frac{dy}{dt}+\frac{\partial^2r}{\partial x\partial z}\frac{dx}{dt}\frac{dz}{dt}+\frac{\partial^2r}{\partial y \partial z}\frac{dy}{dt}\frac{dz}{dt}\right]$$ $$=\frac{\dot{x}^2(y^2+z^2)+\dot{y}(x^2+y^2)+\dot{z}^2(x^2+y^2)-2xy\dot{x}\dot{y}-2xz\dot{x}\dot{z}-2yz\dot{y}\dot{z}}{(x^2+y^2+z^2)^{3/2}}$$ However I am not sure If I have any way of checking this online. I have not written out all of the equations because it is quite long.

Answer 1

I used to calculate related problems.

This is related to the arc length higher derivative in differential geometry.

\begin{align*} \dfrac{\mathrm{d}s}{\mathrm{d}t} =\sqrt{\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 +\left(\dfrac{\mathrm{d}y}{\mathrm{d}t}\right)^2 +\left(\dfrac{\mathrm{d}z}{\mathrm{d}t}\right)^2} &=\left|\dfrac{\mathrm{d}\large\boldsymbol{r}}{\mathrm{d}t}\right| =\boxed{\,\vphantom{\dfrac{+}{}}\Big|{\large\boldsymbol{r}}'\Big|\,}\\ \dfrac{\mathrm{d}^2s}{\mathrm{d}t^2} =\dfrac{\dfrac{\mathrm{d}x}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + \dfrac{\mathrm{d}y}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + \dfrac{\mathrm{d}z}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^2z}{\mathrm{d}t^2} }{\sqrt{\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 +\left(\dfrac{\mathrm{d}y}{\mathrm{d}t}\right)^2 +\left(\dfrac{\mathrm{d}z}{\mathrm{d}t}\right)^2}} &=\dfrac{\dfrac{\mathrm{d}\large\boldsymbol{r}}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^2\large\boldsymbol{r}}{\mathrm{d}t^2}}{\left|\dfrac{\mathrm{d}\large\boldsymbol{r}}{\mathrm{d}t}\right|} =\boxed{\,\dfrac{{\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}''}{\Big|{\large\boldsymbol{r}}'\Big|}\,}\\ \end{align*} \begin{align*} \dfrac{\mathrm{d}^3s}{\mathrm{d}t^3} &=\boxed{\,\dfrac{\big|{\large\boldsymbol{r}}'\big|^2\left(\big|{\large\boldsymbol{r}}''\big|^2+{\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}'''\right)-\left({\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}''\right)^2}{\big|{\large\boldsymbol{r}}'\big|^3}\,}\\ \dfrac{\mathrm{d}^4s}{\mathrm{d}t^4} &=\boxed{\,\dfrac{3\left({\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}''\right)^3-3\big|{\large\boldsymbol{r}}'\big|^2\left({\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}''\right)\left(\big|{\large\boldsymbol{r}}''\big|^2+{\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}'''\right)+\big|{\large\boldsymbol{r}}'\big|^4\left(3{\large\boldsymbol{r}}''\cdot{\large\boldsymbol{r}}'''+{\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}^{(4)}\right)^2}{\big|{\large\boldsymbol{r}}'\big|^5}\,}\\ \end{align*}

%%%%% %%%%% %%%%% %%%%% %%%%% %%%%% %%%%%

\begin{align*} u\big(t\big) =\sqrt{\Big(x(t)\Big)^2 +\Big(y(t)\Big)^2 +\Big(z(t)\Big)^2} &=\Big|\large\boldsymbol{r}\Big| =\boxed{\,\vphantom{\dfrac{+}{}}\Big|{\large\boldsymbol{r}}\Big|\,}\\ u'\big(t\big) =\dfrac{x\cdot\dfrac{\mathrm{d}x}{\mathrm{d}t}+ y\cdot\dfrac{\mathrm{d}y}{\mathrm{d}t}+ z\cdot\dfrac{\mathrm{d}z}{\mathrm{d}t}}{\sqrt{\Big(x(t)\Big)^2 +\Big(y(t)\Big)^2 +\Big(z(t)\Big)^2}} &=\dfrac{\large\boldsymbol{r}\cdot\dfrac{\mathrm{d}\large\boldsymbol{r}}{\mathrm{d}t}}{\Big|\,\large\boldsymbol{r}\,\Big|} =\boxed{\,\dfrac{{\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}'}{\Big|\,{\large\boldsymbol{r}}\,\Big|}\,}\\ \end{align*}

\begin{align*} u''\big(t\big) &=\boxed{\,\dfrac{\big|\,{\large\boldsymbol{r}}\,\big|^2\left(\big|{\large\boldsymbol{r}}'\big|^2+{\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}''\right)-\left({\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}'\right)^2}{\big|\,{\large\boldsymbol{r}}\,\big|^3}\,}\\ \\ u'''\big(t\big)&=\boxed{\,\dfrac{3\left({\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}'\right)^3-3\big|{\large\boldsymbol{r}}\big|^2\left({\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}'\right)\left(\big|{\large\boldsymbol{r}}'\big|^2+{\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}''\right)+\big|\,{\large\boldsymbol{r}}\,\big|^4\left(3{\large\boldsymbol{r}}'\cdot{\large\boldsymbol{r}}''+{\large\boldsymbol{r}}\cdot{\large\boldsymbol{r}}'''\right)^2}{\big|{\large\boldsymbol{r}}\big|^5}\,}\\ \end{align*}

Answer 2

I agree that this is a quite tedious task if we use brute force.

Starting from $$r=\sqrt{x^2+y^2+z^2}$$ where $x=x(t)$, $y=y(t)$ and $z=z(t)$ we have (this part is easy) $$r'=\frac{x x'+y y'+z z'}{\sqrt{x^2+y^2+z^2}}=\color{red}{\frac{x x'+y y'+z z'}{r}}$$ So $$r''=\frac{(x x''+x'^2)+(y y''+y'^2)+(z z''+z'^2)}{r}-\frac{r' \left(x x'+y y'+z z'\right)}{r^2}$$ $$r''=\frac{(x x''+x'^2)+(y y''+y'^2)+(z z''+z'^2)}{r}-\frac{r'^2} r$$

All the trick was to replace the denominator of $r'$ by $r$.